Lesson 6: Theorems About Roots of Polynomial Equations

Property

For a polynomial $P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$ with integer coefficients, any rational root $\frac{p}{q}$ (in lowest terms) must satisfy:

• $p$ divides the constant term $a_0$
• $q$ divides the leading coefficient $a_n$

Examples

Property

Every polynomial of degree $n \geq 1$ has exactly $n$ complex roots (counting multiplicities).

If $P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$ where $n \geq 1$ and $a_n \neq 0$, then $P(x) = 0$ has exactly $n$ solutions in the complex number system.

Property

Synthetic division is a shorthand method for polynomial division. Synthetic division only works when the divisor is of the form $x - c$.
To perform synthetic division, write the coefficients of the dividend and the value $c$ from the divisor. Bring down the first coefficient, multiply by $c$, add to the next coefficient, and repeat. The last number in the result is the remainder.

Examples

• To divide $x^3 + 4x^2 - 5x - 14$ by $x - 2$, use $c=2$ in synthetic division. The coefficients are $1, 4, -5, -14$. The resulting coefficients are $1, 6, 7$ with a remainder of $0$. The quotient is $x^2 + 6x + 7$.
• To divide $3x^3 + 8x^2 + 5x - 7$ by $x + 2$, use $c=-2$. The coefficients are $3, 8, 5, -7$. The process yields a quotient of $3x^2 + 2x + 1$ and a remainder of $-9$.
• To divide $y^4 - 8y^2 + 16$ by $y + 2$, use $c=-2$ and a placeholder for the $y^3$ and $y$ terms (coefficients $1, 0, -8, 0, 16$). The quotient is $y^3 - 2y^2 - 4y + 8$ with a remainder of $0$.

Explanation

Synthetic division is a fast-track version of long division that gets rid of the variables. It's a cleaner and quicker process, but remember its major limitation: it only works when you are dividing by a simple binomial like $(x-5)$ or $(x+2)$.

Property

If a polynomial has rational coefficients and $a + b\sqrt{c}$ is a root (where $a$ and $b$ are rational, $c$ is not a perfect square), then $a - b\sqrt{c}$ is also a root.

Examples